1. Tardigrade
  2. Question
  3. Mathematics
  4. If n is an integer greater than 1 , then a- n C1(a-1)+ n C2(a-2)- ldots+(-1)n(a-n) is equal to :

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Q. If $n$ is an integer greater than 1 , then $a-{ }^{n} C_{1}(a-1)+{ }^{n} C_{2}(a-2)-\ldots+(-1)^{n}(a-n)$ is equal to :

Bihar CECEBihar CECE 2003

Solution:

$a-{ }^{n} C_{1}(a-1)+{ }^{n} C_{2}(a-2)-$
$\ldots+(-1)^{n}(a-n)$
$=a\left({ }^{n} C_{0}-{ }^{n} C_{1}+{ }^{n} C_{2}-{ }^{n} C_{3}+\ldots +(-1)^{n n} C_{n}\right)$
$+\left({ }^{n} C_{1}-2{ }^{n} C_{2}+3{ }^{n} C_{3}-\ldots+(-1)^{n+1} n{ }^{n} C_{n}\right)$ ...(i)
As we know
$(1-x)^{n}={ }^{n} C_{0}-x{ }^{n} C_{1}+x^{2 n} C_{2}-x^{3 n} C_{3}+$
$\ldots+(-1)^{n} x^{n n} C_{n}$ ...(ii)
Put $x=1$, we get
$0={ }^{n} C_{0}-{ }^{n} C_{1}+{ }^{n} C_{2}-{ }^{n} C_{3}+\ldots+(-1)^{n n} C_{n}$
On differentiating Eq. (ii) w.r.t. $x$, we get
$n(1-x)^{n-1}=-{ }^{n} C_{1}+2 x{ }^{n} C_{2}-3 x^{2 n} C_{3}+\ldots$
$+(-1)^{n} n x^{n-1}{ }^{n} C_{n}$
Put $x=1$
$0=-{ }^{n} C_{1}+2^{n} C_{2}-3^{n} C_{3}+\ldots .+(-1)^{n-1} n{ }^{n} C_{n}$
From Eq. (i)
$a-{ }^{n} C_{1}(a-1)+{ }^{n} C_{2}(a-2)-\ldots+(-1)^{n}(a-n)$
$=a(0)+0=0$